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I heard a theorem in differential geometry course.

State of the theorem is

"There is no closed (regular) surface having only negative Gaussian curvature."

I tried to prove the theorem using Gauss-Bonnet theorem, but coudn't have any progress.

How can I get proof?

+) I guess that above theorem is also true when the word 'negative Gaussian curvature' is replaced with 'nonpositive Gaussian curvature'. Is this right?

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    $\begingroup$ It depends on your definition of a surface. $\endgroup$ – Moishe Kohan Jun 8 '16 at 13:28
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Actually there are such surfaces in every genus $g=2,3,4,\ldots$ and in fact loads of them.

The hypothesis that seems to be missing is that the surface is embedded in Euclidean $3$-space. Then indeed you can't have nonpositive curvature. The simplest argument to prove this is to use compactness to find a point on the surface at maximal distance from the origin. Then you can show that Gaussian curvature must be positive at such a point.

Here the Gauss-Bonnet theorem you mentioned does not help because as I mentioned such surfaces do exist if you allow higher-dimensional Euclidean spaces.

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